Calculate the horse power of a man who can chew ice at the rate of `60g min.^(-1)`. Given `J = 4.2 J cal^(-1)` and altent heat of ice `= 80 cal g^(-1), 1hp = 746W`.

Power of a man who can chew 0.3 kg ice in one minute is (in cal/s).

The entropy change in melting 1g of ice at 0^@C . (Latent heat of ice is 80 cal g^(-1) )

1 of a steam at 100^@C melts how much ice at 0^@C ? ( Latent heat of ice=80 cal/g and latent heat of steam =540 cal/g )

75 g of water at 100^@ C is added to 20 g of ice at -15^@C . Calculate the resulting temperature, give that latent heat of ice= 80 cal g^(-1)C^(-1) and specific heat of ice=0.5 cal g^(-1) C^(-1) .

A refrigerator converts 100 g of water at 20^@ C to ice at - 10^@ C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g^(-1) K^(-1) , specific latent heat of ice is 336 J g^(-1) and the specific heat capacity of ice is 2.1 J g^(-1) K^(-1)

What is the horse power applied by a person while chewing 100g of ice per minute ? Given J = 4.2 joule per calorie.

If a ball of 80 kg mass hits an ice cube and temperature of ball is 100 .^(@)C , then how much ice converted into water ? (Specific heat of ball is 0.2 cal g^(-1) , Latent heat of ice = 80 cal g^(-1) )